Heston Model: Calibration and Simulation

Hard·28 min read·Interactive lab

Derivatives PricingStochastic VolatilityHeston ModelFourier PricingCalibration

Quick Quiz

1. The Feller condition for the Heston variance process $dv_t = \kappa(\theta - v_t)\,dt + \xi\sqrt{v_t}\,dW_t^{(2)}$ is $2\kappa\theta > \xi^2$ . What is the implication when this condition is violated?

The SDE is ill-posed and has no solutionThe variance process can reach zero with positive probabilityThe characteristic function of the log-price does not existThe correlation

\rho

must equal zero

2. In the Heston characteristic function $\varphi_T(u) = \exp(iurT + A(u,T) + B(u,T)v_0)$ , the functions $A$ and $B$ satisfy ODEs because the Heston model is:

Complete: every contingent claim can be replicatedAffine: the generator applied to

e^{A+Bv}

separates into terms linear in

v

Markovian: the process

(S_t, v_t)

is a diffusionGaussian: the log-price is normally distributed

3. The full-truncation Euler scheme for the Heston variance process sets $v_t^+ = \max(v_t, 0)$ and uses $v_t^+$ in the drift and diffusion coefficients. What is the key advantage over the 'absorption' scheme (which sets $v_{t+\Delta t} = \max(v_{t+\Delta t}, 0)$ after each step)?

Full-truncation produces smaller variance in the Monte Carlo estimatorFull-truncation allows the variance to remain negative and recover naturally, reducing the upward bias in varianceFull-truncation is faster because it avoids taking the maximum at every stepFull-truncation satisfies the Feller condition exactly

4. In the Levenberg-Marquardt calibration of the Heston model, the parameters $\kappa$ and $\theta$ are individually well-identified from a standard implied vol surface — their individual values can be estimated precisely even when $\kappa\theta$ is constrained.

truefalse

5. Why is the Albrecher et al. (2007) formulation of the Heston characteristic function preferred over the original Heston (1993) formulation for long maturities?

Albrecher's formulation converges faster in the Fourier integralThe original Heston formulation evaluates the characteristic function on the wrong Riemann sheet due to branch-cut discontinuitiesAlbrecher's formulation does not require the Feller conditionThe original Heston characteristic function is only valid for

T < 1

year

6. In the Lewis (2001) formula for the Heston call price, what integration contour is used and why does it avoid the need for a dampening parameter $\alpha$ ?

The real axis; no dampening needed because the Heston characteristic function decays rapidlyThe contour

\mathrm{Im}(u) = -1/2

; the payoff function has a simple Fourier transform along this stripThe contour

\mathrm{Im}(u) = +1

; this avoids the pole at

u = 0

The imaginary axis; the characteristic function is real-valued along it

←Lab